The reaction of CH_3OH (g) to give HCN (g), NH_3 (g), and O_2 (g) requires 164 k

Question

The reaction of CH_3OH (g) to give HCN (g), NH_3 (g), and O_2 (g) requires 164 kj/mol of CH_3 OH (g). A Write a balanced chemical equation for this reaction. Include the phases of all species in the reaction. 2 CH_3 OH (g) + 2N_2 (g) rightarrow 2HCN (g) + 2NH_3 (g) + O_2 (g) Tip: Plus signs (+) can be typed from the keyboard. reaction arrows can be found in the Tools menu of the answer module. Include phases in the chemical equation. Click on the (aq) button in the tools to select the phase to add. Phases should not be subscripted. Use the left and rightarrow keys, or click with the mouse, so move the cursor into or out of parentheses or a subscript in the module. B. Should the thermal energy involved be written as a reactant or as a product? Product Reactant C. How much that is involved in the reaction of 60.0 g of CH_3OH (g) with excess N_2 (g) to give HCN (g) and NH_3 (g) in this reaction?

Explanation / Answer

a) The balanced chemical equation is
2CH3OH (g) + 2N2 (g) -------->  2HCN (g) + 2NH3 (g) +O2 (g)

b) As the energy is required for this reaction this means energy is absorbed so the energy involved is added as the reactant

c) 1 mole of CH3OH requires 164 KJ of energy
   i.e 32g (molar mass ) of CH3OH requires 164 KJ of energy
Applying unitary method :-

1g of CH3OH requires (164/32) KJ of energy
Thus, 60g of CH3OH requires (164/32) x 60 KJ of energy = 307.5 KJ of energy

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